3.1452 \(\int (a+b \cos (c+d x)) (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^{\frac{7}{2}}(c+d x) \, dx\)

Optimal. Leaf size=192 \[ \frac{2 \sin (c+d x) \sqrt{\sec (c+d x)} (3 a A+5 a C+5 b B)}{5 d}+\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) (a B+A b+3 b C)}{3 d}-\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) (3 a A+5 a C+5 b B)}{5 d}+\frac{2 (a B+A b) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{3 d}+\frac{2 a A \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{5 d} \]

[Out]

(-2*(3*a*A + 5*b*B + 5*a*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (2*(A*b +
a*B + 3*b*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) + (2*(3*a*A + 5*b*B + 5*a
*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(5*d) + (2*(A*b + a*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3*d) + (2*a*A*Se
c[c + d*x]^(5/2)*Sin[c + d*x])/(5*d)

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Rubi [A]  time = 0.34606, antiderivative size = 192, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {4221, 3031, 3021, 2748, 2636, 2639, 2641} \[ \frac{2 \sin (c+d x) \sqrt{\sec (c+d x)} (3 a A+5 a C+5 b B)}{5 d}+\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) (a B+A b+3 b C)}{3 d}-\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) (3 a A+5 a C+5 b B)}{5 d}+\frac{2 (a B+A b) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{3 d}+\frac{2 a A \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^(7/2),x]

[Out]

(-2*(3*a*A + 5*b*B + 5*a*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (2*(A*b +
a*B + 3*b*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) + (2*(3*a*A + 5*b*B + 5*a
*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(5*d) + (2*(A*b + a*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3*d) + (2*a*A*Se
c[c + d*x]^(5/2)*Sin[c + d*x])/(5*d)

Rule 4221

Int[(u_)*((c_.)*sec[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Dist[(c*Sec[a + b*x])^m*(c*Cos[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Cos[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSineIntegrandQ[u,
x]

Rule 3031

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e
_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((b*c - a*d)*(A*b^2 - a*b*B + a^2*C)*
Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b^2*f*(m + 1)*(a^2 - b^2)), x] - Dist[1/(b^2*(m + 1)*(a^2 - b^2)),
Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b
^2*d*(m + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1))))*Sin[e + f*x] - b*C*d*(m +
1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && Ne
Q[a^2 - b^2, 0] && LtQ[m, -1]

Rule 3021

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m +
1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*
C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2636

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1))/(b*d*(n +
1)), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[2*n]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rubi steps

\begin{align*} \int (a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac{7}{2}}(c+d x) \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{(a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac{7}{2}}(c+d x)} \, dx\\ &=\frac{2 a A \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d}-\frac{1}{5} \left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{-\frac{5}{2} (A b+a B)-\frac{1}{2} (3 a A+5 b B+5 a C) \cos (c+d x)-\frac{5}{2} b C \cos ^2(c+d x)}{\cos ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 (A b+a B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac{2 a A \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d}-\frac{1}{15} \left (4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{-\frac{3}{4} (3 a A+5 b B+5 a C)-\frac{5}{4} (A b+a B+3 b C) \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 (A b+a B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac{2 a A \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac{1}{5} \left ((3 a A+5 b B+5 a C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\cos ^{\frac{3}{2}}(c+d x)} \, dx+\frac{1}{3} \left ((A b+a B+3 b C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 (A b+a B+3 b C) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{3 d}+\frac{2 (3 a A+5 b B+5 a C) \sqrt{\sec (c+d x)} \sin (c+d x)}{5 d}+\frac{2 (A b+a B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac{2 a A \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d}-\frac{1}{5} \left ((3 a A+5 b B+5 a C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=-\frac{2 (3 a A+5 b B+5 a C) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{5 d}+\frac{2 (A b+a B+3 b C) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{3 d}+\frac{2 (3 a A+5 b B+5 a C) \sqrt{\sec (c+d x)} \sin (c+d x)}{5 d}+\frac{2 (A b+a B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac{2 a A \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 1.223, size = 149, normalized size = 0.78 \[ \frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \left (5 F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) (a B+A b+3 b C)-3 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) (3 a A+5 a C+5 b B)+\frac{\sin (c+d x) (3 (\cos (2 (c+d x)) (3 a A+5 a C+5 b B)+5 a (A+C)+5 b B)+10 (a B+A b) \cos (c+d x))}{2 \cos ^{\frac{5}{2}}(c+d x)}\right )}{15 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^(7/2),x]

[Out]

(2*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-3*(3*a*A + 5*b*B + 5*a*C)*EllipticE[(c + d*x)/2, 2] + 5*(A*b + a*B
+ 3*b*C)*EllipticF[(c + d*x)/2, 2] + ((10*(A*b + a*B)*Cos[c + d*x] + 3*(5*b*B + 5*a*(A + C) + (3*a*A + 5*b*B +
5*a*C)*Cos[2*(c + d*x)]))*Sin[c + d*x])/(2*Cos[c + d*x]^(5/2))))/(15*d)

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Maple [B]  time = 4.082, size = 742, normalized size = 3.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2),x)

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*C*b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x
+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+
2*(A*b+B*a)*(-1/6*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(-1/2+cos(1/2*d*x+1/
2*c)^2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*
d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))-2/5*a*A/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*
c)^4+6*sin(1/2*d*x+1/2*c)^2-1)/sin(1/2*d*x+1/2*c)^2*(12*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2
*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^4-24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-
12*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2
*d*x+1/2*c)^2+24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+3*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+
1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)-8*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c))*(-2*sin(1/2*d*x+1/
2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+2*(B*b+C*a)*(-(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)
*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+2*(-2*sin(1/2*d*x+
1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2)/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*
d*x+1/2*c)^2-1))/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )}{\left (b \cos \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{\frac{7}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)*sec(d*x + c)^(7/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (C b \cos \left (d x + c\right )^{3} +{\left (C a + B b\right )} \cos \left (d x + c\right )^{2} + A a +{\left (B a + A b\right )} \cos \left (d x + c\right )\right )} \sec \left (d x + c\right )^{\frac{7}{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

integral((C*b*cos(d*x + c)^3 + (C*a + B*b)*cos(d*x + c)^2 + A*a + (B*a + A*b)*cos(d*x + c))*sec(d*x + c)^(7/2)
, x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**(7/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )}{\left (b \cos \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{\frac{7}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)*sec(d*x + c)^(7/2), x)